In a car dealership there are 5 models that get displayed in a line in the front of the parking lot for prime viewing. The dealership sells 15 different models. In how many ways can the 5 models be displayed

360360

Step-by-step explanation:

We have the following information:

Number of ways of choosing 5 car models from 15 different models = 15C5

Number of arranging above 5 models = 5!

Therefore, the total number of displayong 5 models would be:

15C5 * 5!

nCr = n! / (r! * (n-r)!)

we replace:

 15! / (5! * (15-5)!) * 5! = 15! / 10! = 360360

So there are a total of 360 360 ways to display the 5 car models.

x+y=90

x=30-y=x

step-by-step explanation:

if you understand you will get the answer

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12